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scott1328 wrote:

newolder wrote:

Evolving wrote:Pi is defined as the ratio between the circumference and the diameter of a circle, and is as such "accurate" by definition. That is the number that we use in trigonometry and in areas of physics that make use of the concept ("angular" concepts like angular momentum). If we want a decimal representation, we can get arbitrarily close, but it's impossible to represent any irrational number exactly as a decimal.

The same applies to any irrational number, such as the square root of two, square root of five etc.

Is the square root of each prime number irrational? I don't know a) the answer (but I think it's a yes) and b) how to prove such a thing.

Happy π day! (It's like a caek day but irrational.)

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Date: 10/08/97 at 17:00:17
From: Terry Dobbins
Subject: Irrational numbers

My question is: Will all square roots of positive numbers that
are not perfect squares be irrational numbers?

I am a new teacher and this was asked of me by another teacher.
I think that it is a true statement but I can't prove it.

Thanks for the help.


Date: 10/08/97 at 18:18:08
From: Doctor Tom
Subject: Re: Irrational numbers

Yes. They are all irrational. The proof is similar to the proof that
sqrt(2) is irrational.

In case you haven't seen that, here's how it goes:

Suppose sqrt(2) is rational. Then you can write sqrt(2) as a/b, where
a and b are integers, and the fraction is reduced to lowest terms.

So a^2/b^2 = 2 so a^2 = 2*b^2. So a is even. Since it's even, write
a = 2*c.  (2c)^2 = 2*b^2 or 4c^2 = 2b^2 or 2c^2 = b^2, so b is also
even. But then you didn't reduce a/b to lowest terms since they both
have a factor of 2.

To show that sqrt(p) is irrational where p is a prime number, the same
approach works, except instead of saying "a is even," you'll be saying
"a is a multiple of p."  The proof goes the same way, except that you
find that a and b are both multiples of p, and hence your original
fraction wasn't reduced as you said it was.

For an arbitrary number n that's not a perfect square, you can factor
it as follows:

n = p1^n1*p2^n2*p3^n3*... for a finite number of terms. At least one
of the n1, n2, n3, ... must be odd, or n is a perfect square.
Suppose n1 is the one that's odd. If n1 is 1, just go through the same
proof above and show that the a and b in your a/b are multiples of p1. 
If n is odd and bigger than one, write your a/b as a*p1^((n1-1)/2)/b. 
That'll get rid of the part of the product of primes that's a perfect
factor of p1.

To make this concrete, suppose I want to show that 216 does not have a
rational square root.

  216 = 2^3*3^3.

If 216 has a rational square root, it will be  2*sqrt(216/2^2)
= 2*sqrt(54), so let sqrt(216) = 2*a/b, reduced to lowest terms. 
Then 4a^2/b^2 = 54, so 2a^2/b^2 = 27, or 2a^2 = 27b^2, so b must be
even. There's already a contradiction.

-Doctor Tom,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/   


Date: 10/08/97 at 18:31:06
From: Doctor Wallace
Subject: Re: Irrational numbers

Dear Terry,

The answer is yes, all non-perfect square square roots are irrational. 
Remember that a rational number is one that can be expressed as the
ratio of 2 integers. If you look in our archives, you'll find a proof
for the fact that the square root of 2 is irrational. Search on the
terms irrational and square root of 2. I won't repeat the details
here, except to say that the proof involves assuming that the square
root of 2  IS  rational, and working to a contradiction. The proof is
simple and elegant. If I remember correctly, there is also a proof in
the archives for the square root of 3.

As to a general proof that ALL non-perfect square square roots are
irrational, I'm not sure. I know that one exists, though. Perhaps it
is accomplished through extension of the two proofs I mentioned. 

I hope this helps.  Don't hesitate to write back if you have more
questions.

-Doctor Wallace,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/